Now, we take a point P.
On the line OP we choose an axis u.
The number t is a value of the angle from the xaxis to the uaxis.
The number r is such that P = r.U
The numbers r and t define unambiguous the point P.
We say that (r,t) is a pair of polar coordinates of P.
One point P has many pairs of polar coordinates. If (r,t) is a pair of polar coordinates, (r, t + 2.k.pi) is also a pair of polar coordinates and additionally ( r, t + (2.k+1).pi ) is a pair of polar coordinates too.
Examples
Figure 1:
Figure 2:
polar coordinates of P are (1.4 , 3.6) or (1.4 , 0.46 ) or (1.4 , 5.8) ...
Figure 3:
polar coordinates of P are (3 , 5.7) or (3 , 0.58) or (3 , 2.56 ) ...
The polar coordinates of the pole O are by definition (0,t) with t perfectly arbitrary.
According to the previous definition, the cartesian coordinates of U are
(cos t, sin t).
Since P = r.U, the cartesian coordinates of p are (r.cos t, r.sin t).
The transformation formulas are x = r.cos t, y = r.sin t
Figure 1: the cartesian coordinates of P are (1.91, 0.59)
Figure 2: the cartesian coordinates of P are (1.25, 0.62)
Figure 3: the cartesian coordinates of P are (2.50, 1.65)
The cartesian coordinates of a point P are (x,y).
Choose the uaxis such that r > 0. then
P = r U => P^{2} = r^{2} U^{2} => x^{2} + y^{2} = r^{2} r = sqrt(x^{2} + y^{2}) Now, choose a tvalue such that x = r.cos t and y = r.sin tIn that way we have a pair of polar coordinates (r,t) of P.
Exercise: Transform the thee pairs of cartesian coordinates of P above back to polar coordinates.
Each point P of that curve has at least one pair of polar coordinates who satisfy the equation. Note that, in general, not all pairs of polar coordinates of P are solutions of the equation.
Note that one curve can have different polar equations.
Exercise 1:
Exercise 2:
Exercise 4:
Given the curve with polar equation r = 2.tan(t).sin(t).
Point A is on the curve and OA = 1/sqrt(3). 0 < t < pi/2.
Find t.
Suppose a curve c has polar equation r = f(t).
t + n = a => n = a  t tan(a)  tan(t) => tan(n) =  1 + tan(a).tan(t) Say the variable point P has cartesian coordinates (x,y). Then we know that dy y tan(a) =  and tan(t) =  dx x Thus, dy y    dx x x dy  y dx tan(n) =  =  dy y x dx + y dy 1 +  .  dx x From x = r cos(t) and y = r sin(t) we have dx = dr.cos(t)  r.sin(t) . dt dy = dr.sin(t) + r.cos(t) . dt From this we calculate x dy  y dx = r^{2} dt From x^{2} + y^{2} = r^{2} we find 2 x dx + 2 y dy = 2 r dr or x dx + y dy = r dr Now, we can simplify tan(n) r^{2} dt r tan(n) =  =  = r dr dr/dt r tan(n) =  r'
The following formula gives the direction of the curve c at any point P.
Here r = f(t) is the equation of the curve and r' stands for (dr/dt) = f'(t).
From r and r' you can calculate the direction n (see figure above ) at
each point.
r tan(n) =  r' 
Exercise 2:
We take the curve with polar equation r = 1/(1cos(t)). It is a parabola with cartesian
equation y^{2} = 2x+1. Draw the parabola. Choose point P on the parabola with polar coordinates (2 , pi/3).
Draw OP and the tangent line in point P. Draw the angle n as in the previous figure.
Calculate the angle with the formula tan(n) = r/r' and verify your result on your figure.
cot(n) = r'/r = (ln(r))' Now (ln(r))' = k for all t.First take k = 0. Then ln(r) = constant and r is constant. The curves are the circles with midpoint in the origin. Now, take k not 0. Then
(ln(r))' = k dln(r) <=>  = k dt <=> ln(r) = k t + constant we denote the constant as ln(m) <=> ln(r) = k t + ln(m) <=> r = m e^{kt}These isogonal curves are called the logarithmic spirals or the Bernouilli spirals.
Exercise:
Give different values to the parameters m and k and plot some spirals.
Let t vary so you can see a few turns.
Calculate k so that the constant angle is 60 degrees. (answer: k = 1/sqrt(3) )
the tangent line is parallel to the polar axis <=> t + n = k.pi <=> tan(t) =  tan(n) r <=> tan(t) =   r'
The following formula gives the tvalues of all points of a curve r=f(t),
where the tangent line is parallel to the polaraxis. The value r' stands for (dr/dt) = f'(t). To calculate the tvalues, you have to solve this trigonometric equation. r tan(t) =   r' 
the tangent line is orthogonal to the polar axis <=> t + n = pi/2 + k.pi <=> tan(t) = cot(n) r' <=> tan(t) =  r
The following formula gives the tvalues of all points of a curve r=f(t), where
the tangent line is orthogonal to the polaraxis. The value r' stands for (dr/dt) = f'(t). To calculate the tvalues, you have to solve this trigonometric equation. r' tan(t) =  r 
2 pi  t_{o} t_{o} t_{o} cos() = cos(pi  ) =  cos  =  r_{o} 2 2 2From this, we see that (r_{o}, 2pit_{o}) is a solution of r = cos(t/2).
r' =  (1/2) sin(t/2)The tangent line is parallel to the polaraxis if and only if
tan(t) =  r/r' <=> tan(t) = 2 cot(t/2) To solve this equation we let u = tan(t/2) , then 2 u 2 <=>  =  1  u^{2} u <=> ... ___ ___ V 2 V 2 <=> tan(t/2) =  or tan(t/2) =   2 2This gives the value t = 1.23 in [0,pi]. Then r = 0.816
P on C => F(x_{o},y_{o}) = 0 => F(r_{o}.cos(t_{o}), r_{o}.sin(t_{o})) = 0 => P on C'
P on C' => F(r_{o}.cos(t_{o}), r_{o}.sin(t_{o})) = 0 => F(x_{o},y_{o}) = 0 => P on K'
Suppose a curve C has a cartesian equation F(x,y) = 0.
If we replace x by r.cos(t) and y by r.sin(t),
then we have a polar equation
F(r.cos(t),r.sin(t))=0 of the curve C. 
r^{2} cos^{2}(t) + r^{2} sin^{2}(t)  4r cos(t) = 0 <=> r^{2}  4r cos(t) = 0 <=> ( r= 0 or r  4 cos(t) = 0 )r= 0 is the equation of the pole itself. So, the pole belongs to the curve.
Example 2:
We want to find a polar equation of the curve
y^{2} (1 + x)  x^{2} (1  x) = 0 We replace x by r.cos(t) and y by r.sin(t). r^{2} sin^{2}(t) ( 1 + r cos(t))  r^{2} cos^{2}(t) (1  r cos(t)) = 0 <=> r^{2} ( sin^{2}(t)  cos^{2}(t) + r cos(t) sin^{2}(t) + r cos^{3}(t)) = 0 <=> r = 0 or  cos(2t) + r cos(t) = 0 <=> r = 0 or cos(2t) = r cos(t) cos(2t) <=> r = 0 or r =  cos(t)Since r = cos(2t)/cos(t) contains the pole, we omit the first part r=0.
Exercise: Show that the hyperbola x y = 2 has a polar equation r^{2} sin(2t) = 4.
Suppose a curve C has a polar equation G(r,t) = 0.
If that equation can be transformed to an equation 
r^{2} cos^{2} (t) + r^{2} sin^{2} (t)  4 = 0 <=> x^{2} + y^{2} = 4
Example 2
Suppose a curve C has a polar equation r = 3 sin(t).
For t = 0, we find r = 0.Thus, the pole is part of C and we don't add a point to C if we
multiply both sides by r.
C has polar equation r^{2} = 3 r sin(t).
The cartesian equation of C is x^{2} +y^{2} + 3y = 0. C is a circle.
Example 3
Suppose a curve C has a polar equation r = 1 + 2.cos(t).
If we choose t such that cos(t) = 0.5, then r = 0.
Thus, the pole is part of C and we don't add a point to C if we
multiply both sides of r = 1 + 2.cos(t) by r.
C has a polar equation r^{2} = r + 2.r.cos(t) <=> r = r^{2}  2.r.cos(t)
Now take the curve C' : r = ( r^{2}  2.r.cos(t)) .
We'll show that each point of C belongs to the curve C'.
Therefore we take an arbitrary point P(r_{o},t_{o}) on C.
Since P is on C we have:
r_{o} = r_{o}^{2}  2.r_{o}.cos(t_{o}) (*)But the same point P has polar coordinates ( r_{o}, t_{o} + pi ). So, we can write P(r_{o},t_{o} + pi). We'll show that P is on C'.
P( r_{o}, t_{o} + pi) is on C' <=>  r_{o} = ( r_{o}^{2} + 2.r_{o}.cos(t_{o} + pi) ) <=> r_{o} = r_{o}^{2}  2.r_{o}.cos(t_{o}) and this is true because of (*)In the same way we can show that each point of C' is on C. (exercise)
Hence, we can write
C has a polar equation r = (+1 or 1).( r^{2}  2.r.cos(t)) <=> C has a polar equation r^{2} = (r^{2}  2 r cos(t))^{2} <=> C has a cartesian equation x^{2} + y^{2} = ( x^{2} + y^{2} 2 x )^{2}
Cardioid Cissoid of Diocles Cochleoid Conchoid Trifolium Folium Folium of Descartes Hyperbolic Spiral Lemniscate of Bernoulli Right Strophoid etc...
Go and look at
u(r_{1} , t_{1} ) and v(r_{2} , t_{2} ) In the corresponding cartesian coordinate system the two vectors have cartesian coordinates u(x_{1} , y_{1} ) and v(x_{2} , y_{2} ) with x_{1} = r_{1} cos(t_{1} ) x_{2} = r_{2} cos(t_{2} ) y_{1} = r_{1} sin(t_{1} ) y_{2} = r_{2} sin(t_{2} ) The dot product u.v = x_{1} x_{2} + y_{1} y_{2} = r_{1} cos(t_{1} ) . r_{2} cos(t_{2} ) + r_{1} sin(t_{1} ).r_{2} sin(t_{2} ) = r_{1} r_{2} (cos(t_{1} )cos(t_{2} ) + sin(t_{1})sin(t_{2} )) = r_{1} r_{2} cos(t_{1}  t_{2} )
The dot product of two vectors u (r_{1} , t_{1} ) and v (r_{2} , t_{2} ) is u . v = r_{1} r_{2} cos(t_{1}  t_{2} ) 
Find the values of t such that the dot product of OP and OQ is equal to 2.
Afterwards, take such value of t and draw the vectors OP and OQ.
Check geometrically whether the scalar product is 2.
A point P(r,t) is on line d <=> PN is orthogonal to ON <=> PN . ON = 0 <=> (N  P). N = 0 <=> N.N  P.N = 0 <=> r_{o}.r_{o}.cos(0)  r.r_{o}.cos(t  t_{o}) = 0 Since r_{o} and cos(t  t_{o}) are not 0 r_{o} <=> r =  cos(t  t_{o})
The pole is not on a line d. The line l, through the pole and perpendicular to the line d, intersects line d in point N(r_{o},t_{o}). The equation of line d is r_{o} r =  cos(t  t_{o}) 
Exercise 2:
r = 12/( 6*sin(t)  4 cos(t) ) is a polar equation of a curve c.Find the cartesian equation of c.
A circle with the pole as center and radius R has a polar equation r = R 
P(r,t) is on the circle <=>  CP  = R <=>  CP ^{2} = R^{2} <=> (P  C)^{2} = R^{2} <=> P^{2}  2 P C + C^{2} = R^{2} <=> r^{2}  2 r r_{o} cos(t  t_{o}) + r_{o}^{2} = R^{2}
A circle, with C(r_{o},t_{o}) as center and R as radius, has has a polar equation
r^{2}  2 r r_{o} cos(t  t_{o}) + r_{o}^{2} = R^{2} 
r^{2}  2 r R cos(t) = 0 <=> r = 0 or r = 2R cos(t) <=> r = 2R cos(t) (since this curve already contains the pole for t = pi/2)
A circle, with C(R,0) as center and R as radius, has has a polar equation
r = 2R cos(t) 
In cartesian coordinates we find the coordinates of the common
points by solving the system of the two equations.
But if we solve the system of the polar equations we find only one
point with polar coordinates (2,1).
Now we'll show
Similarly,
Say curve c' has a polar equation F'(r,t)=0.
Call V' the set of all solutions of the equation F'(r,t)=0.
Then curve c' is the set of all points corresponding with the set V'.
With each element of V', there is one and only one point of curve c'.
With an element of the intersection of V and V' corresponds a common point of c and c', BUT it is possible that V contains a solution (r_{o},t_{o}) and that V' contains a different solution (r_{1},t_{1}) and that both solutions correspond with a common point of c and c'. Then (r_{o},t_{o}) and (r_{1},t_{1}) are different polar coordinates of that same point.
In the example above we have
(2,1) is a solution of t = 1 and this gives a point of the line l.
(2,1+pi) is a solution of r = 2 and this gives a point of the circle c.
Although these solutions are different, the corresponding point is a
common point of the line and the circle.
We say that an equation F(r,t)=0 of a curve c has property (P) if and only if ALL polar coordinates of EACH point of c are solutions of F(r,t)=0Corollary:
Say curve c has a polar equation F(r,t)=0 with the property (P), and curve c' has a polar equation F'(r,t)=0. Now the set of solutions of the system of the two polar equations contains the coordinates of ALL the common points of the two curves and the problem has disappeared!
If at least one of the two equations has the property (P), then we can find all intersection points of the two curves by solving the system of the two equations. (The pole sometimes escapes, see below)
We know that r_{o} r =  (1) cos(t  t_{o})is the equation of a line d.
Example:
A line d contains point D(3, pi/4) and stands perpendicular to line OD.
A circle c has center O and radius = 5.
The intersection points are the solutions of the system
/ r = 5  3  r =  \ cos(t  pi/4) We have cos(t  pi/4) = 3/5 cos(t  pi/4) = cos(0.927) t  pi/4 = 0.927 or t  pi/4 = 0.927 This gives the points with polar coordinates (5, 1.71) and (5, 0.14)
r^{2}  2 r r_{o} cos(t  to) + r_{o}^{2} = R^{2} (2)is the equation of a circle.
This is also true for the circle with the equation r = 2R cos(t)
r^{2} = R^{2}is an equation of the same circle and this equation has property (P).
Example:
We calculate the intersection points of the curves with equation
r = cos(t/2) and r = 1/2 We have to solve the system /  r^{2} = 1/4   r = cos(t/2) \ <=> /  r = 1/2 or r = 1/2   r = cos(t/2) \ <=> / /  r = 1/2  r =  1/2  or   cos(t/2)=1/2  cos(t/2)=  1/2 \ \ <=> ..... We find four solutions: (1/2, 2 pi / 3) or ( 1/2, 4 pi / 3) or (1/2,  2 pi / 3) or ( 1/2,  4 pi / 3)The circle and the curve have four common points.
t = t_{o} + k.piis an equation of the same line and this equation has property (P).
Example: We calculate the intersection points of the curves with equation
r = 4 cos(t) and r = 4 sin(t)Since the first equation has property (P), the intersection points are the solutions of the system
r = 4 cos(t) r = 4 sin(t)The coordinates of the pole are not a solution of that system but the pole IS an intersection point of the curves.
To calculate the common points of curve c with equation F(r,t)=0 and c' with equation F'(r,t)=0, it is sufficient to solve the system of the equations F(r,t)=0 and F'(r,t)=0 if at least one of the equations has the property (P). But even then you have to investigate separately if the pole is a common point. 
Exercise:
Find the intersection points of the circle r = 2 sin(t) and the cardioid r = 1+cos(t).
Given:
5 K has equation r =  ( 1  2 sin(t) + 3 cos(t) ) 2 L has equation r =  ( 2 cos(t) + sin(t) )Calculate

Then K has equation
5 r =  ( 1  e cos(tt_{o}) )And this is the equation of a conic section.
2 r =  A.cos(tt_{o})This is the equation of a line
Since the equation of L has the property (P), the intersection points of K and L are the solutions of the system
5 r =  ( 1  2 sin(t) + 3 cos(t) ) 2 r =  ( 2 cos(t) + sin(t) )From these equations we have
5 2  =  ( 1  2 sin(t) + 3 cos(t) ) ( 2 cos(t) + sin(t) ) <=> .... <=> 4 cos(t) + 9 sin(t) = 2and with the method from here we solve this equation.
<=> .... <=> t_{1} = 2.518876 and t_{2} = 0.2137328 The corresponding values of r are : r_{1} = 1.92058 and r_{2} = 1.14785
To convert the equation of K to a cartesian equation, we appeal on the properties of Polar equation of a conic section.
The conic section K has polar equations
5 5 r =  and r =  ( 1  2 sin(t) + 3 cos(t) ) ( 1 + 2 sin(t)  3 cos(t) ) <=> r  2r sin(t) + 3 r cos(t) = 5 and r + 2r sin(t)  3 r cos(t) = 5 <=> r = 5 + 2 r sin(t)  3 r cos(t) and r = ( 5 + 2 r sin(t)  3 r cos(t)) <=> r^{2} = ( 5 + 2 r sin(t)  3 r cos(t))^{2} <=> x^{2} + y^{2} = (5 + 2 y  3 x)^{2} <=> 8x^{2}  12 xy + 3 y^{2}  30 x + 20 y + 25 = 0For the line L, the transformation is easy.
L has equation 2 r cos(t) + r sin(t) = 2 <=> 2 x + y = 2To calculate the intersection points in cartesian coordinates, we solve the system
8x^{2}  12 xy + 3 y^{2}  30 x + 20 y + 25 = 0 2 x + y = 2With simple algebra we find
( 1.1217 ; 0.24347) and (1.560 ; 1.120 )It is easy to verify that these points are the same points as above.
c with equation F(r,t)=0 (1) and c' with equation F(r, t  alpha)=0 (2) Now we have P(r_{o},t_{o}) is a solution of (1) <=> P(r_{o},t_{o} + alpha) is a solution of (2)This means that we obtain the curve c' by rotating the curve c by an angle alpha (center of rotation is O).
Example 1:
If we rotate the circle r = 4cos(t) by an angle of pi/2 radians, the new circle has equation
r = 4 cos(t  pi/2) <=> r = 4 sin(t)
We rotate the curve c with equation F(r,t)=0 by an angle alpha. The new curve has equation F(r, t  alpha)=0. 
We rotate the curve c with equation r = sin(2t) by an angle of pi radians. The new curve c' has equation
r = sin(2(tpi)) <=> r = sin(2t  2pi) <=> r = sin(2t)We find that c and c' have the same equation. This means that c is invariant for a rotation over pi radians.
Suppose a curve c has polar equation r = f(t).
Say N = l S and T = m S (this defines the numbers l and m) NP.PT = 0 (P  N).(T  P) = 0 P.T  P.P  N.T + N.P = 0 0  P.P  N.T + 0 = 0 P.P =  N.T r.r = l.m
P = N + NP P.B = N.B + NP.B r U.B = l S.B + 0 r cos(n) = l sin(n) l = r cot(n) 1 dr dr dr l = r . =  (l is the visualization of  ) r dt dt dt
1 l 1 dr d 1  =   =  . =  () m r^{2} r^{2} dt dt r
1 d 1  =  () = g'(t) <=> m = 1/ g'(t) m dt r So, m_{o} = 1/ g'(t_{o})From m_{o} we denote point T and then we draw the asymptote.
The equation of the asymptote is
m_{o} r =  cos(t(t_{o}+pi/2)) m_{o} <=> r =  sin(t t_{o})
To calculate an asymptotes of the curve c with polar equation r = f(t),
we take four steps:

Take the curve c with equation r = 3/(1+2cos(t)).
Then g(t) = (1+2cos(t))/3 and g'(t) = 2sin(t)/3
g(t) = 0 for t_{o} = 2pi/3 and t_{1} = 2pi/3
These values are the directions of the asymptotes.
The corresponding values of m are
m_{o} =  sqrt(3) and m_{1} = sqrt(3).
The asymptotes are
 sqrt(3) sqrt(3) r =  and r =  sin(t  2pi/3) sin(t + 2pi/3)In a graph this gives a hyperbola and his two asymptotes.
Exercise: Find the asymptote of the curve r = 2 tan(t) . sin(t)
A curve has polar equation r = f(t). Consider the colored part bounded by OAB. One can prove that the area of this part is equal to
t2 /  1   r^{2} dt  2 / t1
Since the curve is symmetrical relative to the polar axis, we first calculate the area of the upper half. The area of this part is:
pi /  1   4 a^{2} (1 + cos(t))^{2} dt  2 / 0 pi /  =2a^{2}  (1 + 2 cos(t) + cos^{2}(t)) dt  / 0 now: cos^{2}(t) = (1/2) (1 + cos(2t)) pi = 2 a^{2} [ t + 2 sin(t) + (1/2)( t + (1/2) sin(2t) ] 0 = 2 a^{2} (3/2) pi = 3 pi a^{2}The area of the whole cardioid is 6 pi a^{2}
Exercise 1:
The polar equation r = sin(3t) gives a threebladed curve. Plot the curve.
Find the area of 1 blade.
Exercise 2:
Calculate the area of the part of the circle r = 4 cos(t) located outside the circle r = 1.5
Exercise 3:
Calculate the area of the region bounded by the smallest loop of the curve r = 1 + 2 cos(t).
A curve has a polar equation r=f(t). Consider the arc AB.
t2 /   sqrt( r^{2} + r'^{2}) dt  / t1
r' =  sin(t) r^{2} + r'^{2} = 1 + 2 cos(t) + cos^{2}(t) + sin^{2}(t) = 2 (1 + cos(t)) = 4 cos^{2}(t/2) The length a half cardioid is then pi /  pi  2 cos(t/2) dt = 2 [2 sin(t/2)] = 4  0 / 0The total length is 8.